# Defining a partial order on $A\times B$, given partial orders on $A$ and on $B$

Let $(A,\preceq_A)$ and $(B,\preceq_B)$ be partially ordered sets. Define $C = A \times B$ and define the relation $\preccurlyeq$ on $C$ to be $(a,b) \preccurlyeq$ $(a',b')$ if and only if $(a\preceq_A a') \wedge (b\preceq_B b')$. I'm to prove here that $\preccurlyeq$ is a partial order on $C$. I know that to prove it is indeed a partial order, it must be reflexive, transitive and anti-symmetric. So I think for reflexivity you'd have $(a,a) \preccurlyeq$ $(a',a')$ where $C = A \times A$, which is pretty obvious. I'm not sure how one would prove the other parts, though.

• For reflexivity you have to show that $(a,b)\preccurlyeq(a,b)$. Aug 23, 2014 at 5:52
• What is the problem in following the definition? Aug 23, 2014 at 6:04

$[(a,b) \preccurlyeq$ $(a',b')] \wedge [(a',b') \preccurlyeq$ $(a'',b'')] \iff$ $[(a\preceq_A a') \wedge (b\preceq_B b')] \wedge[(a'\preceq_A a'') \wedge (b'\preceq_B b'')] \iff$ $[(a\preceq_A a') \wedge (a'\preceq_A a'')] \wedge[(b\preceq_B b') \wedge (b'\preceq_B b'')] \Rightarrow$
$[(a\preceq_A a'') \wedge (b\preceq_B b'')] \iff$ $[(a,b) \preccurlyeq$ $(a'',b'')]$